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3 years ago

How much water should we add to 5 gallons of 18% acid solution to dilute it to a concentration of 10%?

Mathematics

1 answer:

KIM [24]3 years ago

5 0

4 gallons of water is needed to dilute to a concentration of 10%.

Step-by-step explanation:

It is given that,

The 5 gallons of water contains 18% of acid solution.

Now, we are asked to find the gallons of water needed to dilute it to a concentration of 10% acid solution.

We obtain the equation by stating that the amount of acid 10% in the water must dilute up to 18%.

For this, x amount of water should be added.

Therefore, the equation is framed as,

⇒ 0.1 (x+5) = 5 (0.18)

⇒ 0.1x + 0.5 = 0.9

⇒ 0.1x = 0.9 - 0.5

⇒ x = 0.4 / 0.1

⇒ x = 4

∴ 4 gallons of water is needed to dilute to a concentration of 10%.

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A building 84 ft high casts a 46 ft shadow. How tall is a building that casts a 110 ft shadow at the same time of the day? (to t

Bond [772]

The ratio of building heights will be the same as the ratio of shadow lengths.
(taller building)/(shorter building) = (longer shadow)/(shorter shadow)
(taller building)/(84 ft) = (110 ft)/(46 ft) . . . . . . fill in the given numbers
(taller building) = (84 ft)*(110/46) ≈ 200.9 ft . . multiply by 84 ft, evaluate

The appropriate selection is
D) 200.9 ft

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3 years ago

Read 2 more answers

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

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2 years ago

What’s the next three after? 18,36,72,144,..

IgorC [24]

Answer: 288, 576, 1152

Step-by-step explanation: What you do here is that you multiply every number by two. You start by doing 18 times 2, which equals 36. Then you do 36 times 2, which equals 72. You keep on doing this until you get the answers needed.

Hope this helped!! :)

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3 years ago

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Swimming pool pete's pools installs rectangular pools that are all 10 feet wide. the length can very from 10 feet to 30 feet. th

AleksandrR [38]

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3 years ago

Suppose a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%. A cell phon

leva [86]

Answer:

We conclude that the actual percentage of households is equal to 30%.

Step-by-step explanation:

We are given that a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%.

Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

Let p = proportion of households that have three cell phones NOT known.

So, Null Hypothesis, $H_0$ : p = 30% {means that the actual percentage of households is equal to 30%}

Alternate Hypothesis, $H_A$ : p $\neq$ 30% {means that the actual percentage of households different from 30%}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of households having three cell phones = $\frac{43}{150}$ = 0.29

n = sample of households = 150

So, the test statistics = $\frac{0.29-0.30}{\sqrt{\frac{0.30(1-0.30)}{150} } }$

= -0.27

The value of z test statistic is -0.27.

Also, P-value of the test statistics is given by;

P-value = P(Z < -0.27) = 1 - P(Z $\leq$ 0.27)

= 1 - 0.6064 = 0.3936

Now, at 1% significance level the z table gives critical value of -2.58 and 2.58 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual percentage of households is equal to 30%.

4 0

3 years ago